codeforces 607B Zuma 区间dp

题目如下：

B. Zuma

time limit per test
2 seconds

memory limit per test
512 megabytes

input
standard input

output
standard output

Genos recently installed the game Zuma on his phone. In Zuma there exists a line of n gemstones, the i-th
of which has color ci.
The goal of the game is to destroy all the gemstones in the line as quickly as possible.

In one second, Genos is able to choose exactly one continuous substring of colored gemstones that is a palindrome and remove it from the line. After the substring is removed, the remaining gemstones shift to form a solid line again. What is the minimum number
of seconds needed to destroy the entire line?

Let us remind, that the string (or substring) is called palindrome, if it reads same backwards or forward. In our case this means the color of the first gemstone is equal to the color of the last
one, the color of the second gemstone is equal to the color of the next to last and so on.

Input

The first line of input contains a single integer n (1 ≤ n ≤ 500) —
the number of gemstones.

The second line contains n space-separated integers, the i-th
of which is ci (1 ≤ ci ≤ n) —
the color of the i-th gemstone in a line.

Output

Print a single integer — the minimum number of seconds needed to destroy the entire line.

Sample test(s)

input

3
1 2 1

output

1

input

3
1 2 3

output

3

input
7
1 4 4 2 3 2 1


output
2


Note

In the first sample, Genos can destroy the entire line in one second.

In the second sample, Genos can only destroy one gemstone at a time, so destroying three gemstones takes three seconds.

In the third sample, to achieve the optimal time of two seconds, destroy palindrome 4 4 first and then destroy palindrome 12 3 2 1.

我的想法：

    其实一开始自己完全没有头绪，主要是自己想去找回文串然后删掉构造新串，而这种方法无疑是指数级的，而且也没有什么高效的找回文串的方法。在网上看到高人的博客后恍然大悟，其实完全可以没必要去考虑整个回文串，只需要注意到一个事实：如果a[l]==a[r]&&l<r那么消去子串[l,r]的时间就等于消去子串[l-1,r+1]的时间。这样一来，我们就可以把问题分解，进而想到用记忆化搜索。

    具体来说，对于一个串，我们只需要考虑与a[l]相同的元素，以其为划分划分成两部分即可。注意特殊判定一下a[l]==a[r]的部分就可以了。

代码如下：

#include <iostream>

using namespace
std;

const int maxn =
502;

int n;

int a[maxn];

int f[maxn][maxn];

int dp(int l,
int r){

    if (l>=r) return
1;

    if (f[l][r]>0) {//如果状态已存在，则返回（记忆化搜索）

        return f[l][r];

    }

    f[l][r] = 1+dp(l+1,r);//相当于下面的i=l的情况，给f[l][r]一个初值而已

    for (int i=l+1; i<=r; i++) {

        if (a[l] ==
a[i]) {

            f[l][r] =
min(f[l][r],
dp(l+1,i-1)+dp(i+1,r));

        }

    }

    if(a[l] ==
a[r]) f[l][r] =
min(f[l][r],
dp(l+1,r-1));//特殊判定，因为此时不用划分

    return f[l][r];

}

int main(int argc,
const char * argv[]) {

    cin >> n;

    for (int i=0; i<n; i++) {

        cin >> a[i];

    }

    cout << dp(0,n-1);

    return 0;

}